Question #124660
Verify that the point P(a cos θ, b sin θ) lies on the ellipse
x
2
a
2
+
y
2
b
2
= 1,
where a and b are the semi-major and semi-minor axes respectively of the ellipse . Find the
gradient of the tangent to the curve at P and show that the equation of the normal at P is
ax sin θ − by cos θ = (a
2 − b
2
) sin θ cos θ.
If P is not on the axes and if the normal at P passes through the point B(0, b), Show that
a
2 > 2b
2
. If further, the tangent at P meets the y-axis at Q, show that
|BQ| =
a
2
b
2
.
1
Expert's answer
2020-07-01T19:41:41-0400

If the point P(acosθ,bsinθ)P(acos\theta,bsin\theta) satisfies the equation x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 , then it lies on the given ellipse .


Now x2a2+y2b2=(acosθ)2a2+(bsinθ)2b2\frac{x^2}{a^2}+\frac{y^2}{b^2} = \frac{(acos\theta)^2}{a^2}+\frac{(bsin\theta)^2}{b^2}

=a2cos2θa2+b2sin2θb2= \frac{a^2cos^2\theta}{a^2}+\frac{b^2sin^2\theta}{b^2}

=cos2θ+sin2θ= cos^2\theta+sin^2\theta

=1.=1.

\therefore point P(acosθ,bsinθ)P(acos\theta,bsin\theta) lies on the given ellipse. (PROVED).


Now the equation of the tangent at the point P(acosθ,bsinθ)P(acos\theta,bsin\theta) is given by

xacosθa2+ybsinθb2=1or, xcosθa+ysinθb=1or, bxcosθ+aysinθab=1or, bxcosθ+aysinθ=abor, aysinθ=bxcosθ+abor, y=bxcosθasinθ+abasinθor, y=(batanθ)x+bcosecθ.\frac{x*acos\theta}{a^2}+\frac{y*bsin\theta}{b^2}=1\\or, \space \frac{xcos\theta}{a}+\frac{ysin\theta}{b}=1\\or, \space \frac{bxcos\theta+aysin\theta}{ab}=1\\or, \space bxcos\theta+aysin\theta=ab\\or, \space aysin\theta=-bxcos\theta+ab\\or, \space y=-\frac{bxcos\theta}{asin\theta}+\frac{ab}{asin\theta}\\or, \space y=-(\frac{b}{a}tan\theta)x +bcosec\theta.


\therefore Gradient of the tangent at the point P(acosθ,bsinθ)P(acos\theta,bsin\theta) is (batanθ)-(\frac{b}{a}tan\theta) . Answer.


Equation of Normal: We know that normal is perpendicular to the tangent..


so if equation of tangent is bxcosθ+aysinθ=abbxcos\theta+aysin\theta=ab then equation of normal is of the form


axsinθbycosθ=k, where k is constant.axsin\theta-bycos\theta=k, \space where \space k \space is \space constant.


Now as this normal is passing through P(acosθ,bsinθ)P(acos\theta,bsin\theta) ...by putting this coordinate into the equation of normal we get the value of kk .


aacosθsinθbbsinθcosθ=kor, k=a2sinθcosθb2sinθcosθor, k=(a2b2)sinθcosθ\therefore a*acos\theta*sin\theta-b*bsin\theta*cos\theta=k\\or, \space k=a^2sin\theta cos\theta-b^2sin\theta cos\theta\\or, \space k= (a^2-b^2)sin\theta cos\theta


\therefore The equation of normal at the point P(acosθ,bsinθ)P(acos\theta,bsin\theta) is given by

axsinθbycosθ=(a2b2)sinθcosθ.axsin\theta-bycos\theta= (a^2-b^2)sin\theta cos\theta. PROVED.


Now if the equation of normal is passing through B(0,b) then we have

a0sinθbbcosθ=(a2b2)sinθcosθor, b2cosθ=(a2b2)sinθcosθ.or, b2=(a2b2)sinθor, sinθ=b2a2b2a*0*sin\theta-b*b*cos\theta=(a^2-b^2)sin\theta cos\theta\\or, \space -b^2cos\theta=(a^2-b^2)sin\theta cos\theta.\\ or, \space -b^2=(a^2-b^2)sin\theta \\or, \space sin\theta = \frac{-b^2}{a^2-b^2}


Now as we know 1sinθ,-1\leq sin\theta,\\

1b2a2b2or,1b2a2b2or,a2b2b2or,a22b2\therefore -1\leq \frac{-b^2}{a^2-b^2}\\or, 1\geq \frac{b^2}{a^2-b^2}\\ or, a^2-b^2 \geq b^2\\or, a^2 \geq2b^2 PROVED.


Now tangent at P(acosθ,bsinθ)P(acos\theta,bsin\theta) which is bxcosθ+aysinθ=abbxcos\theta+aysin\theta=ab , meets the Y axis at Q.

putting x=0 ,we get

0+aysinθ=abor, ysinθ=bor, y=bsinθ.0+aysin\theta=ab\\or, \space ysin\theta=b\\or, \space y=\frac{ b}{sin\theta}.\\


\therefore The coordinate of Q is (0,bsinθ)(0,\frac{b}{sin\theta}) .


BQ=(bsinθb)BQ=(bb2a2b2b)BQ= \bigg(\frac{b}{sin\theta}-b\bigg)\\BQ=\bigg(\frac{b}{\frac{-b^2}{a^2-b^2}}-b\bigg)

BQ=b(b2a2b21)BQ=b(b2a2b2b2)BQ=a2bBQ= b(\frac{b^2-a^2}{b^2}-1)\\BQ= b*(\frac{b^2-a^2-b^2}{b^2})\\BQ=\frac{-a^2}{b}

BQ=a2b\therefore |BQ|= \frac{a^2}{b} PROVED.


After solving question i found that BQ is a2b\frac{a^2}{b} but in a given question it was given as a2b2\frac{a^2}{b^2} .. This is a mistake.








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